ammonium nitrate sodium hydroxide reaction
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A nitrate test is a chemical test used to determine the presence of nitrate ion in solution. Testing for the presence of nitrate via wet chemistry is generally difficult compared with testing for other anions, as almost all nitrates are soluble in water. In contrast, many common ions give insoluble salts, e.g. halides precipitate with silver, and sulfates precipitate with barium.
The nitrate anion is an oxidizer, and many tests for the nitrate anion are based on this property. Unfortunately, other oxidants present in the analyte may interfere and give erroneous results.
Brown ring test
A common nitrate test, known as the brown ring test can be performed by adding iron(II) sulfate to a solution of a nitrate, then slowly adding concentrated sulfuric acid such that the sulfuric acid forms a layer above the aqueous solution. A brown ring will form at the juction of the two layers, indicating the presence of the nitrate ion. Note that the presence of nitrite ions will interfere with this test.
The overall reaction is the reduction of the nitrate ion by iron(II) which is oxidised to iron(III) and formation of a nitrosyl complex.
- NO3- + 3Fe2+ + 4H+→ 3Fe3+ + NO + 2H2O
- [Fe(H2O)6]2+ + NO → [Fe(H2O)5(NO)]2+
Devarda's alloy (Cu/Al/Zn) is a reducing agent. When reacted with nitrate in sodium hydroxide solution, ammonia is liberated. The ammonia formed may be detected by its characteristic odor, and by moist red litmus — very few gases other than ammonia evolved from wet chemistry are alkaline.
- 3 NO|3|- + 8 Al + 5 OH|- + 18 H|2|O â†’ 3 NH|3 + 8 [Al(OH)|4|]|-
The aluminium does the reducing in this reaction.
Diphenylamine may be used as a wet chemical test for the presence of the nitrate ion. In this test, a solution of diphenylamine and ammonium chloride in sulfuric acid is used. In the presence of nitrates, diphenylamine is oxidized, giving a blue coloration. This reaction has been used to test for organic nitrates as well, and has found use in gunshot residue kits detecting nitroglycerine and nitrocellulose.
Other oxidants such as chlorate, bromate, etc. interfere by similarly oxidizing diphenylamine. They may be removed by reduction with sodium sulfite. Where nitrite is present, a false negative result may be observed due to sulfite reducing nitrate in the presence of nitrite.
The diphenylamine test may be selective for nitrate by reducing nitrite with sodium azide, prior to treatment with sodium sulfite. Other derivatives have been reported as well.
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Answers:Phenolphthalein reacts with bases to give a pink colour. This is because a base will take 2 protons off phenolphthalein, and the resulting ion is pink- 2NH4OH + C20H14O4 --> 2H2O + C20H12O4-- 2NH4+ This is an acid-base reaction. With ammonium hydroxide and acetic acid, they will react together to neutralise each other, creating a neutral solution, leaving phenolphthalein as the uncoloured, neutral molecule- NH4OH + C20H14O4 + CH3COOH --> H2O + CH3COONH4 + C20H14O4 This is a neutralisation reaction.
Answers:Sulphuric acid is a diprotic acid (that means it has 2 H+ that can react) therefore t undergoes two step neutralisation when reacted with NaOH 1.) NaOH + H2SO4 NaHSO4 + H2O . Please note that this reaction does not give a neutral solution, That is at pH = 7.0. The solution is still strongly acidic. But I have answered in this way in view of your second question. The salt is called sodium hydrogen sulphate 2.) NaHSO4 + NaOH Na2SO4 + H2O This is the final end point, the solution will be neutral and the quantity of H2SO4 consumed in total will be twice that consumed in the first reaction. The saltformed is called sodium sulphate 3 .1 CaCO3 +2 HNO3 Ca(NO3)2 + CO2 + H2O React 1 mol CaCO3 with 2 mol HNO3 to give Ca(NO3)2 , carbon dioxide and water 3.2 The two steps are a) Ca(OH)2 + 2CO2 Ca(HCO3)2 That is calcium hydrogen carbonate which is formed as a first step. b) This will break down on drying to give the calcium carbonate Ca(HCO3)2(aq) CO2(g) + H2O(l) + CaCO3(s).
Answers:Total ionic equation: HC2H3O2 + Na+ + OH- -> H2O + Na+ + C2H3O2- The acetic acid is a weak acid, so it must be written as a molecule. Sodium hydroxide is a strong base, meaning that 100% of its ions break up, so it has to be written separately. Water is a polar covalent compound, so it has to be written as a molecule. What we have left are sodium and acetate ions floating around in solution. Sodium (Na+) is the spectator ion, so if we remove this ion from both sides of the equation, we have the net ionic equation: HC2H3O2 + OH- -> H2O + C2H3O2-